Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Section 5.4 - Product-to-Sum and Sum-to-Product Formulas - Exercise Set - Page 691: 56

Answer

The required solution is $\frac{1}{2}\left[ \sin \left( \alpha +\beta \right)+\sin \left( \alpha -\beta \right) \right]=\sin \alpha \sin \beta $.

Work Step by Step

We know that the sum and difference formulas are $\sin \left( \alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta $ and $\sin \left( \alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta $. Add the left and right sides of these two identities: $\begin{align} & \sin \left( \alpha +\beta \right)+\sin \left( \alpha -\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta +\sin \alpha \cos \beta -\cos \alpha \sin \beta \\ & =2\sin \alpha \cos \beta \end{align}$ Now, multiply both sides by $\frac{1}{2}$. $\begin{align} & \frac{1}{2}\left[ \sin \left( \alpha +\beta \right)+\sin \left( \alpha -\beta \right) \right]=\frac{1}{2}\cdot 2\sin \alpha \cos \beta \\ & =\sin \alpha \cos \beta \end{align}$ Therefore, the product-to-sum formula is derived, which is $\sin \alpha \cos \beta =\frac{1}{2}\left[ \sin \left( \alpha +\beta \right)+\sin \left( \alpha -\beta \right) \right]$. Hence, by adding the left and right sides of the identities, the product-to-sum formula is derived for $\sin \alpha \cos \beta $.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.